Answer:
They eat Himalayan blue sheep, boar, deers, wild mountain goats, pikas etc.
Explanation:
Two point charges, initially 1 cm apart, are moved to a distance of 3 cm apart. By what
factor do the resulting electrostatic forces between them change?
A lamp is 15% efficient. How much electrical energy must be supplied to the lamp each second if it produces 30 J of light energy per second? *WILL MARK BRAINLIEST*
200 Joules should be supplied in the circuit.
Efficiency = output/inuptHere the output is 30J and input is missing
[tex]\sf \dfrac{30}{I} * 100 = 15[/tex]
multiplied with 100, to covert it into percentage
[tex]\sf 3000 = 15(I)[/tex]
[tex]\sf I = \dfrac{3000 }{15}[/tex]
[tex]\sf I = 200J[/tex]
Let that be X
Efficiency is 15% means the lamp produces 15J of light energy when 100J energy applied.[tex]\\ \rm\Rrightarrow \dfrac{30}{x}(100)=15[/tex]
[tex]\\ \rm\Rrightarrow 3000=15x[/tex]
[tex]\\ \rm\Rrightarrow x=200J[/tex]
A 2.0-kg block slides on a rough horizontal surface. A force (magnitude P = 4.0 N) acting parallel to the surface is applied to the block. The magnitude of the block's acceleration is 1.2 m/s2. If P is increased to 5.0 N, determine the magnitude of the block's
When the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².
Frictional force between the block and the horizontal surfaceThe frictional force between the block and the horizontal surface is determined by applying Newton's law;
∑F = ma
F - Ff = ma
Ff = F - ma
Ff = 4 - 2(1.2)
Ff = 4 - 2.4
Ff = 1.6 N
When the applied force increases to 5 N, the magnitude of the block's acceleration is calculated as follows;
F - Ff = ma
5 - 1.6 = 2a
3.4 = 2a
a = 3.4/2
a = 1.7 m/s²
Thus, when the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².
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I need help with this please
Answer:
Rahman was a little bit 5AM but I think he was just try and the guy was the one who just got a a??ze and you know how are you to me that you
An aircraft starts at rest and is accelerated for 9.2 seconds, at which time the aircraft launches. If the distance traveled from the starting point to the launch point was 428.8 m, what was the launch velocity of the aircraft in m/s? Assume the acceleration is constant. (Hint: multi-step problem).
Explanation:
Since the acceleration is constant, we use the big four equations.
[tex]x = x _{i} + vt[/tex]
[tex]x = \frac{1}{2} a {t}^{2} + v _{i}t + x _{i}[/tex]
[tex]v {}^{2} = (v _{i}) {}^{2} + 2a(x - x _{i})[/tex]
[tex] \frac{1}{2} (v + v _{i})t = x - x _{i}[/tex]
Next. identity the variables.
We know
t=9.2
change in position or (x- x_i) is 428.8
Initial Velocity is 0.
We need to solve for the final velocity, so we use the fourth equation.
[tex] \frac{1}{2} (v - v _{i})t = x - x _{i}[/tex]
Subsitue
[tex] \frac{1}{2} (v - 0)(9.2) = 428.8[/tex]
[tex] \frac{1}{2} (v)9.2 = 428.8[/tex]
[tex]v = 93.22 \frac{m}{s} [/tex]
A rod is 4.0 m long and has a square cross-section that is 1.5 cm on each side. An ohmmeter measures 0.040 Ω across its ends. What is the resistivity of the material from which this rod is made?
Answer:
2.25 x 10^-6 ohm - m 2.25 x 10^-4 ohm - cm
Explanation:
rho = R A/l
R = .040
A = .015 * .015 m^2
l = 4 rho =
What are your ideas about how humans can reduce the amount of carbon dioxide and methane they add to the atmosphere?
Answer:
Find the heavy processes of CO2 and methane.
Explanation:
Step #1 Find a very reliable energy source.
Step #2 Find the heavy processes of CO2 and methane like in many factories and places with high carbon emissions.
Step #3 Shut down each of the heavy processes even 10 per day would do a lot.
Step #4 Use the renewable energy source as a replacement for the attempts to generate energy using carbon emissions.
Kinetic is calculated using ___
Answer:
Ek= 1/2*m*v^2
Explanation:
kinetic energy= 1/2* mass * square velocity
Question 5 of 10
Which of the following is most likely to happen when energy is transferred to
an object?
O A. The force on the object will decrease.
O B. The object will begin to move.
O C. The object will stay exactly the same.
O D. The mass of the object will increase.
Answer:
B: The object will start moving
Explanation:
If energy is transferred the object will definitely change so it can't be a. If you add energy the object will have more force so it cant be c. The mass of an object can't increase just by giving an object energy so it cant be d
A proton in the nucleus of an atom has an electrical charge of:
neutral
-
+
zero
Answer:
proton is positively charged changechar
Explanation:
Vector approach in solving forces in 3d
Answer:
[tex] \red{ \rule{1pt}{500000pt}}[/tex] [tex] \red{ \rule{1pt}{500000pt}}[/tex]
Explanation:
At a typical bowling alley the distance from the line where the ball is released (foul line) to the first pin is 60ft. Assume it takes 5.0s for the ball to reach the pins after you release it, if it rolls without slipping and has a constant translational speed. Also assume the ball weights 12lb and has a diameter of 8.5 inches.
A- Calculate the rotation rate of the ball in rev/s
B- What is the total kinetic energy in pound-feet? Ignore the finger holes and treat the bowling ball as a uniform sphere
A) The rotation rate of the ball in rev/secs is ; 10.79 rev/secs
B) The total kinetic energy of in pound-feet ignoring the finger holes is : 37.6157 pound-feet
Given data :
mass of ball = 12 Ib
diameter of ball = 8.5 inches
Radius = 4.25 inches = 0.10795 m
Time = 5 secs
distance travelled = 60 ft = 18.288 m
A) Determine the rotation rate of the ballFirst step : calculate the velocity of ball
V = distance / time
= 18.288 / 5 = 3.66 m/s
Next step :
angular velocity ( w ) = V / r
= 3.66 / ( 0.10795 ) = 33.90 rad/sec
convert to rev/sec = 33.90 / π
= 10.79 rev/secs
B) Determine the total kinetic energy
given that ball rolls without slipping
Vcon = Rw
Total K.E = 1/2 IW² + 1/2 mv²
= 7/10 MR²W²
Total K.E = 7/10 * 5.44 * ( 0.10795 )² * 33.90²
= 50.99 ≈ 51 J
convert to pound-feet = 37.6157 pound-feet
Hence we can conclude that The rotation rate of the ball in rev/secs is ; 10.79 rev/secs and The total kinetic energy of in pound-feet ignoring the finger holes is : 37.6157 pound-feet.
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Which has more momentum, a 5000 kg truck moving at 20 km per hour or a 1000 kg truck moving at 50 km per hour?
Answer:
First truck has greater momnetum
Explanation:
momentum = m * v
truck a 5000 * 20 = 10 0000
truck b 1000 * 50 = 5 0000
Match the missions with their objectives.
NEAR
Lunar Prospector
Deep Impact
SETI
mapping the Moon's surface composition,
looking for possible deposits of polar ice,
and examining the Moon's resources,
structure, and origins
first space probe to orbit and perform
an in-depth Investigation of an asteroid
searching for extraterrestrial life in the universe
probing beneath the surface of a comet to
obtain clues about its internal composition
and structure
Matching the missions with their objectives
NEAR -------> First space probe to orbit and perform an in-depth Investigation of an asteroid. Lunar prospector ----- > mapping the Moon's surface composition,looking for possible deposits of polar ice,and examining the Moon's resources,structure, and origins.Deep Impact -------> probing beneath the surface of a comet to obtain clues about its internal composition and structureSETI -----> searching for extraterrestrial life in the universeThe Mission OverviewThe NEAR ( Near Earth Asteroid Rendezvous ) is a mission whose objective is to study origin and nature of objects near the earth such as asteroids and comets.
Lunar prospector was a mission designed to perform an orbit investigation of the moon.
The deep impact mission created by NASA was to probe beneath the comet and discover the structure and composition. The deep impact projected lasted for 9 years in space.
Hence we can conclude that the missions and their objectives are as listed above.
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The maximum kinetic energy of electrons ejected from a sheet of uranium exposed to 250 nm ultraviolet light is 1.33 EV.
What is the work function of the uranium metal? (Planck's constant (h) is 6.626 × 10-34 J∙s, c is 3.0 × 108 m/s, 1 eV = 1.6 × 10-19 J)
Answer: The maximum kinetic energy KEe of ejected electrons (photoelectrons) is given by KEe = hf − BE, where hf is the photon energy and BE is the binding energy (or work function) of the electron to the particular material.
Explanation:
The work function of the uranium metal will be given by the value 7.9*10^-19
What is work function?The work function of a particle is defined as the product of the planks constant to the frequency of the particle.
[tex]\phi=h\times f[/tex]
Now the frequency will be given as
[tex]f=\dfrac{c}{\lambda}[/tex]
So the formula will become
[tex]\phi=\dfrac{hc}{\lambda}[/tex]
h=6.626 × 10-34 J∙s
c=3.0 × 108 m/s
[tex]\lambda[/tex] = 250 nm 250* 10^-9
[tex]\phi=\dfrac{6.626\times 10^{-26}\times3\times 10^{8}}{250\times 10^{-9}}[/tex]
[tex]\phi=7.9\times 10^{-19}[/tex]
Hence the work function of the uranium metal will be given by the value 7.9*10^-19
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A cosmic ray photon is bombarding a massive object to pieces. The detectors indicates
that two fragments, each of mass 0, depart due to such a collision each moving at
the speed of 0.6c at the angle of 600 relative to the photon’s original direction of
motion. What is the energy of the cosmic ray photon in terms 0 and c ?
Answer:
Chapter 1
1. Show that the Lorentz transformation is such that the velocity of a light ray
travelling in the x direction is the same for the observer in the frame S and for
the observer in the frame S
.
Solution: Consider a light ray travelling in the x direction. If the light ray
connects two space–time points {t1, x1} and {t2, x2}, we have
c = x2 − x1
t2 − t1
The speed of light observed in the frame S will be
c = x
2 − x
1
t
2 − t
1
= c
γ ((x2 − x1) − βc(t2 − t1))
γ (c(t2 − t1) − β(x2 − x1))
= c
x2 − x1
t2 − t2
− βc
c − β x2 − x1
t2 − t2
= c
2. What is the mean path before decay for a charged pion with a kinetic energy of
1 GeV?
Solution: The pion has a lifetime 2.6 × 10–8 s and a mass of 139.6 MeV. If the
energy is 1 GeV, the velocity of the pion is 99% of the velocity of light (Eq. 1.4).
The mean path before decay is
= 0.99 c γ τ
= 0.99 c
1000 + 139.6
139.6
2.6 10−8 = 63 m
S. Tavernier, Experimental Techniques in Nuclear and Particle Physics, 271
DOI 10.1007/978-3-642-00829-0, C Springer-Verlag Berlin Heidelberg 2010
272 Solutions to Exercises
3. Show that the relativistic expression for the kinetic energy of a particle (Eq. 1.2)
reduces to the non-relativistic expression if the velocity of the particle is small
compared to the velocity of light.
Solution:
E = Ekinetic + m0c2 = m0c2
1 − (v/c)2
≈ m0c2
(1 − 1/2(v/c)2) ≈ m0c2(1 + 1/2(v/c)
2)
= m0c2 +
1
2
m0v2
4. For a Poisson distribution with average value 16, calculate the probability to
observe 12, 16 and 20 as measured value. Calculate the probability density function for a Gaussian distribution with average value 16 and dispersion 4, for the
values x = 12, 16 and 20. Compare the results.
Solution: For a Poisson distribution P(12) = 0.0829, P(16) = 0.1024, P(20) =
0.0418
For a Gaussian distribution, f(12) = 0.0605, f(16) = 0.0997, f(20) = 0.0605
5. Consider a very short-lived particle of mass M decaying into two long-lived particles 1 and 2. Assume you can measure accurately the energies and momenta of
the two long-lived particles. How will you calculate the mass of the short-lived
particle from the known energies and momenta of the two long-lived objects?
Solution: The mass of the short-lived particle, its energy and its momentum are
related by Eq. (1.1). The energy and momentum of the particle are equal to the
sums of the energy and sums of the momenta of the decay products, therefore
M2c4 = (E1 + E2)
2 − c2(P1 + P2)
2
6. Calculate the order of magnitude of the energy levels in atoms and in nuclei
using the ‘particle in a box’ approximation, Eq. (1.9). Use for the dimension of
the atom 10–10 m and for the dimension of the nucleus 10−15 m.
Solution: Atomic energy levels: ≈40 eV; nuclear energy levels: ≈400 MeV.
7 . Show that in a β− or a β+ decay only a very small fraction of the energy derived
from the mass difference goes to the kinetic energy of the final-state nucleon.
The electron is relativistic; therefore this requires a relativistic calculation! Hint:
the 3-body problem can be reduced to a 2-body problem by considering the
electron–neutrino system as one object with a mass of a few MeV.
Solution. Consider the 2-body decay of some heavy object with mass M into
two objects with masses m1 and m2. The kinetic energy of each of the final-state
particles in the overall centre of mass system is found as follows.
Solutions to Exercises 273
Consider two particles with energy and momentum four vectors p1 and p2.
The symbol pi stands for the four-vector {Ei,cpi}. The energy E appearing in this
expression is the total energy E, i.e. the rest energy mc2 plus the kinetic energy.
The four-vector product (p1.p2) is defined as
(p1.p2) =
(
E1E2 − c2 p1 p2
)
A four-vector product is a Lorentz invariant; this quantity can be evaluated in
any reference frame, and the result is the same. Consider now the quantity
(p1.p2)
m1c2
This is a Lorentz invariant. Evaluating this expression in the rest frame of
particle 1 makes clear that this is the energy of particle 2 seen in the rest frame
of particle 1. This remains true also if one of the particles is in fact a system
of particles, for example the system of the two particles 1 and 2. The energy of
particle 2, seen in the overall centre of mass frame of the particles 1 and 2 is
therefore
E∗
2 = (p1 + p2).p2
(p1 + p2)2
We have the following relations:
(p1 + p2)
2 = M2c4
(p1.p2) = 1
2
(
(p1 + p2)
2 − (p1)
2 − (p2)
2
)
= M2c4 − m2
1c4 − m2
2c4
And therefore finally
E∗
2 = M2c4 + m2
2c4 − m2
1c4
2Mc2
Let us now apply the above relation to the decay
N∗ → N + e− + ¯νe + Q
The symbol Q represents the energy liberated in the reaction. Let us denote
by M∗ the mass of the parent nucleus, by M the mass of the final-state nucleus
and by m the mass of the electron–neutrino system. The kinetic energy of the
nucleus in the final state is given by
274 Solutions to Exercises
Ekin = M∗2c4 + M2c4 − m2c4
2M∗c2 − Mc2
= M∗2c4 + M2c4 − m2c4 − 2M∗c2Mc2
2M∗c2
= (M∗ − M)
2 c4 − m2c4
2M∗c2
=
!
mc2 + Q
The energy of the cosmic ray photon is zero. his means that the photon had insufficient energy to create new particles, and instead, it simply scattered off the massive object.
What is Einstein's energy equation?Einstein's energy equation, also known as the mass-energy equivalence, relates the energy E of an object to its mass m and the speed of light c. The equation is:
E = mc^2
where:
E is the energy of the object in joules (J)
m is the mass of the object in kilograms (kg)
c is the speed of light in meters per second (m/s)
This equation means that mass and energy are interchangeable, and that a small amount of mass can be converted into a large amount of energy. The equation is an important consequence of Einstein's theory of special relativity, and it has been confirmed by numerous experiments, including nuclear reactions and particle accelerators.
Here in the Question,
We can use the conservation of momentum and energy to solve this problem. Since the two fragments have equal mass and are moving in opposite directions at the same speed, we know they have equal and opposite momenta. Therefore, the initial momentum of the photon must also be equal and opposite to the total momentum of the fragments.
Let's call the initial momentum of the photon p and the mass of each fragment m. The total momentum of the fragments is:
p' = 2mv
where v is the speed of each fragment, which we know is 0.6c. Therefore, we can write:
p' = 2m(0.6c) = 1.2mc
By conservation of momentum, we have:
p = -p'
where the negative sign indicates that the photon is moving in the opposite direction to the fragments. Therefore:
p = -1.2mc
Now we can use conservation of energy to relate the photon's energy E to its momentum p:
E^2 = p^2c^2 + m^2c^4
Substituting the expression we found for p, we get:
E^2 = (1.2mc)^2c^2 + m^2c^4
E^2 = 1.44m^2c^4 + m^2c^4
E^2 = 1.45m^2c^4
Solving for E, we get:
E = mc^2 * sqrt(1.45)
Plugging in the values for m and c, we get:
E = (0 * 9.0 × 10^16 kg) * sqrt(1.45) = 0
Therefore, The photon from a cosmic ray has no energy. This indicates that the photon was merely scattered off the large object since it lacked the energy to produce new particles.
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A class of 10 students taking an exam has a power output per student of about 200W. Assume the initial temperature of the room is 20 C and that its dimensions are 6.0 m by 15.0 m by 3.0 m. What is the temperature of the room at the end of 1.0 h if all the energy remains in the air in the room and none is added by an outside source? The specific heat is 837 J/kg C, and its density is about 1.3 x 10^-3 g/cm^3
For a class of 10 students taking an exam has a power output per student of about 200W, the temperature is mathematically given as
dT=25C
What is the temperature of the room at the end of 1.0 h if all the energy remains in the air in the room and none is added by an outside source?Generally, the equation for the temperature is mathematically given as
dT=Q/mc
Where
Q=Pt
Q=2000*3600
Q=7200000J
And m
m=pv
m=1.3*270
m=351kg
Therefore
dT=7200000J/351*837
dT=25C
In conclusion, The temperature of the room at the end of 1.0 h
dT=25C
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A 7500 kg truck is speeding up at 5 m/s down a 1000m on-ramp to the highway.
What is the force exerted on the truck?
Answer:
37500 N
Explanation:
F = ma
F = 7500 (5 m/s^2) = 37500 N
Find the Product or expression of:
(6x+4)2
Answer:
20 or 12
Explanation:
if it is 20 u will have to multiple 6 by 4 and the answer u will multiple it by2 and when it is 12 u will have to add 6 +4 and the answer u will add it to 2
What are some INDIRECT applications of centripetal force in real life?
Answer:
The circular turning of roads
Explanation:
please mark brainliest
Answer:
Explanation:
The circular turning of roads
Driving on Curves
please answer this question
Answer:
The circuits that were made previously were large and bulky, consisting of circuit components like resistor, capacitor, inductor, transistor, diodes, etc., which were connected with copper wires.
Explanation:
I am in confusion with mono electronic circuit and I am not want to made u confused person.
I am really sorry brother.
:-befrank
what are the four things that affect the resistance of a wire? A. length, diameter, material, and temperature B. weight, diameter, material, and temperature C. length, height, material, temperature D. length, weight, material, and temperature
Answer:
A. length, diameter, material, temperature
Find the Product or expression of: (4x-2)2
Answer:
8x-4Explanation:
Distributive property:
A(B+C)=AB+AC
(4x-2)2
Switch sides.
→ 2(4x-2)
Multiply.
⇒ 2*4x=8x
⇒ 2*2=4
Rewrite the problem down.
⇒ = 8x-4
[tex]\Longrightarrow: \boxed{\sf{8x-4}}[/tex]
Hence, the final answer is 8x-4.I hope this helps you! Let me know if my answer is wrong or not.
Answer: 8x - 4
Explanation:
(4x -2)*2
apply distributive property
2* 4x - 2 * 2
Multiply the monomials = 8x
2*4x - 2*= 8x-2*2
calculate =
8x-4
Answer: 8x - 4
When a speeding truck hits a stationary car, the car is deformed and heat is generated. What can you say about the kinetic energy of the system after the collision?
nothing
Explanation:
we can say that it was certainly bad for the shopkeeper obviously and we should not be making questions about the physics behind that accident and should call the cops or 911
Which property of matter is conserved in chemical reactions and shown by balanced equations?
O mass
O volume
O density
O shape
Answer:
Mass.
Explanation:
I took the quiz and got the answer right
determine if a sentence is an argument?
Answer:
The tone really matters and if there are any exclamation marks also.
Explanation:
6. A spring is originally 1.5 m long. It is stretched so that it is 1.8 m long. The spring constant is 50 N/m.
Find the potential energy stored in the spring.
Answer:
PE = 1/2 K X^2
K = 50 N/m
X = .3 m
PE = 1/2 * 50 * .3^2 = 2.25 Joules
write about Archimedes principle
Answer:
Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially, is equal to the weight of the fluid that the body displaces. Archimedes' principle is a law of physics fundamental to fluid mechanics.
Answer:
When a body is immersed fully or partially in a fluid, it experience is an upward force that is equal to the weight of the fluid displaced by it.
Applications:In designing ships and submarines.Lactometers used to determine the purity of sample of milk.Hydrometer is used to determine the density of liquids.
Which statement about the interference behavior of electromagnetic radiation seen in the double-slit test experiment is true? (1 point)
O Waves that make up the radiation collide with each other so that they add together or cancel each other out.
O Particles that make up the radiation collide with each other so that they add together or cancel each other out.
O Particles that make up the radiation collide with each other and scatter randomly.
O Waves that make up the radiation do not interact with each other.
Waves that make up the radiation collide with each other so that they add together or cancel each other out.
What is Interference?This is the phenomenon which occurs when two waves meet while traveling along the same medium.
In the double-slit test experiment, bright and dark fringes was observed which was caused by the superposition of overlapping light waves originating from the two slits which is why option A was chosen as the most appropriate choice.
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A standing wave of the third overtone is induced in a stopped pipe, 3 m long. The speed of sound is 340 m/s. The number of antinodes in the standing wave pattern is:
Answer:
overtone- one over the first
n skips by twos
4 antinodes
500 Hz
Explanation: Hope this helps :)